Difference between revisions of "1951 AHSME Problems/Problem 35"

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==Problem==
 
==Problem==
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If <math> a^{x}= c^{q}= b </math> and <math> c^{y}= a^{z}= d </math>, then
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<math> \textbf{(A)}\ xy = qz\qquad\textbf{(B)}\ \frac{x}{y}=\frac{q}{z}\qquad\textbf{(C)}\ x+y = q+z\qquad\textbf{(D)}\ x-y = q-z </math>
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<math> \textbf{(E)}\ x^{y}= q^{z} </math>
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==Solution==
 
==Solution==
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Try solving both equations for <math>a</math>. Taking the <math>x</math>-th root of both sides in the first equation and the <math>z</math>-the root of both sides in the second gives <math>a=c^{\frac{q}x}</math> and <math>a=c^{\frac{y}z}</math>.
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So <math>\frac{q}x=\frac{y}z</math>. Multiplying both sides by <math>xz</math>, <math>qz=xy</math>. <math>\boxed{\textbf{(A)}}</math>.
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== See Also ==
 
== See Also ==
 
{{AHSME 50p box|year=1951|num-b=34|num-a=36}}  
 
{{AHSME 50p box|year=1951|num-b=34|num-a=36}}  

Latest revision as of 15:28, 19 April 2014

Problem

If $a^{x}= c^{q}= b$ and $c^{y}= a^{z}= d$, then

$\textbf{(A)}\ xy = qz\qquad\textbf{(B)}\ \frac{x}{y}=\frac{q}{z}\qquad\textbf{(C)}\ x+y = q+z\qquad\textbf{(D)}\ x-y = q-z$ $\textbf{(E)}\ x^{y}= q^{z}$

Solution

Try solving both equations for $a$. Taking the $x$-th root of both sides in the first equation and the $z$-the root of both sides in the second gives $a=c^{\frac{q}x}$ and $a=c^{\frac{y}z}$. So $\frac{q}x=\frac{y}z$. Multiplying both sides by $xz$, $qz=xy$. $\boxed{\textbf{(A)}}$.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 34
Followed by
Problem 36
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